Question

A particle moves along the straight line y =D. The speed of the particle is V m/s towards right then find d theta by DT .​

Answer 1

Answer:

$\frac{D \times V}{D^{2} + V^{2} \times t^{2}  }$

Explanation:

tan ∅ = perpendicular/base

From the figure, it is clear that

tan ∅ = x/D

∴ ∅ = $tan^{-1} (\frac{x}{D} )$

At first we will be calculating x;

Since, speed = v;

And , speed = distance / time

speed = x/t

∴ x = Vt

putting  x = Vt in ∅, we get

∴ ∅ = $tan^{-1} (\frac{Vt}{D} )$

Now, we have to differentiate it to get $\frac{d∅}{dt}$

$\frac{d∅}{dt}$ = $\frac{1}{1+Vt^{2} }  \times\frac{V}{D}$

= $\frac{D \times V}{D^{2} + V^{2} \times t^{2}  }$

Answer 2

Answer:

Explanation: ans is

 VCOS ^2THETA